Angles in standard position

1. The problem statement, all variables and given/known data
1) If the point (-4, 2) lies on the terminal arm of an angle θ in standard position, determine the exact value of csc θ.
2) If sec θ = -5/3 and angle θ terminates in quadrant III, which point must be on the terminal side of θ?
3) If cos θ = 5/13, where θ is in quadrant VI, determine the value of cot θ
4) Determine csc θ if (-10, 24) lies on the terminal arm of angle θ in standard position.

2. Relevant equations
I know (cos θ, sin θ)

3. The attempt at a solution

Well for #1, I thought you're supposed to csc(2) since csc θ is 1/sin θ but it's not, and the answer is [tex] \sqrt {5} [/tex] and I have no idea how you get that.

I only know that it's in the second quadrant and 2 is the sin coord. Other than that I have no idea because the book doesn't say how to do these types of questions and I have the test tomorrow. Even if you tell me how to get root 5 for that one question only, I might be able to figure it out.

"In standard position" means that the angle has one side along the positive x-axis. In particular, if the other side passes through the point (-4,2), which has distance [itex]\sqrt{(-4)^2+ 2^2}= \sqrt{20}= 2\sqrt{5}[/itex] from the origin, then it also passes through the point [itex](-2/\sqrt{5}, 1/\sqrt{5})[/itex] which has distance 1 from the origin. I interpret your "2 is the sin coordinate" as meaning that you recognize that the y-coordinate is the one that gives you the sine value for that angle. Of course the x-coordinate is the one that gives you the cosine value at the point where the angle side passes through the unit circle. In other words, knowing that the side passes through (-4, 2) tells you that the sine of the angle is [itex]1/\sqrt{5}[/itex] and the cosine of the angle is [itex]2/\sqrt{5}[/itex]. Now, what is the cosecant of the angle?